Question

How do you solve the equation $ 2{{x}^{2}}+3x-2=0 $ ?

Answer 1

Hint: We are given $ 2{{x}^{2}}+3x-2=0 $, to solve this we learn about Quadratic equation, number of solutions of a quadratic equation. We will learn how to factor the quadratic equation, we will simplify by taking common terms out then we use the zero product rule to get our answer. In the end, we will also learn about quadratic formulas for solving such equations in an easy way and a more speedy way.

Complete step by step answer:
We are given $ 2{{x}^{2}}+3x-2=0 $, we are asked to solve the given problem.
First, we observe that it has a maximum power of ‘2’ so it is a quadratic equation.
Now we should know that a quadratic equation has 2 solutions or we say an equation of power ‘n’ will have an ‘n’ solution.
Now we have to solve the equation $ 2{{x}^{2}}+3x-2=0 $ .
To solve this equation we first take the greatest common factor possibly available to the terms.
As we can see that in $ 2{{x}^{2}}+3x-2=0 $ .
Nothing is common, to solve the equation, we will use the middle term split.
Middle term split method says for any quadratic equation $ 2{{x}^{2}}+3x-2=0 $ .
1) We first product $ a\times c $
2) We try to find numbers such that their sum or difference is ‘b’ while the product is the same as $ a\times c $ .
3) We use that and separate terms.
4) We use zero product rules to get our solution.
Now we have $ 2{{x}^{2}}+3x-2=0 $ .
So, we have $ a=2,b=3,\text{ and }c=-2 $
So, $ a\times c=2\times -2=-4 $
Now we can see that $ 4\times -1=-4 $
And also $ 4-1=3 $
So, we use this to split the middle term ‘b’.
So,
 $ 2{{x}^{2}}+3x-2=0 $ become
 $ 2{{x}^{2}}+\left( 4-1 \right)x-2=0 $
By simplifying, we get –
 $ 2{{x}^{2}}+4x-x-2=0 $
Now, take common from first 2 terms and last 2 terms, we get –
 $ 2x\left( x+2 \right)-1\left( x+2 \right)=0 $
By simplifying further we get –
 $ \left( 2x-1 \right)\left( x+2 \right)=0 $
Now using zero product rules, we have –
 $ 2x-1=0 $ or $ x+2=0 $
So, we get –
 $ 2x=1 $ or $ x+2=0 $
 $ x=\dfrac{1}{2} $ or $ x=-2 $
So, we get $ x=\dfrac{1}{2} $ and $ x=-2 $ are the solution.

Note:
Another way to solve this is we will use Quadratic formula,
Quadratic formula states that for any $ a{{x}^{2}}+bx+c=0 $ solution is given as –
 $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $
For $ 2{{x}^{2}}+3x-2=0 $
We have $ a=2,b=3\text{ and }c=-2 $
So using these, we get –
 $ \dfrac{-3\pm \sqrt{{{3}^{2}}-4\times 2\times \left( -2 \right)}}{2\times 2} $
By simplifying, we get –
 $ x=\dfrac{-3\pm \sqrt{9+16}}{4}=\dfrac{-3\pm \sqrt{25}}{4}=\dfrac{-3\pm 5}{3} $
So, we will get solution from above equation as –
 $ x=\dfrac{-3+5}{4} $ and $ x=\dfrac{-3-5}{4} $
So, simplifying further we get –
 $ x=\dfrac{1}{2} $ and $ x=-2 $ .